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Question Number 85456 by jagoll last updated on 22/Mar/20

∫ (√(csc x )) dx?

$$\int\:\sqrt{\mathrm{csc}\:\mathrm{x}\:}\:\mathrm{dx}? \\ $$

Commented by jagoll last updated on 22/Mar/20

thank sir

$$\mathrm{thank}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 22/Mar/20

sadly this is wrong. t=(√(sin x)) → dx=((2(√(sin x)))/(cos x))dt=((2t)/(√(1−t^4 )))dt ⇒ 2∫(dt/(√(1−t^4 ))) and we can′t solve this

$$\mathrm{sadly}\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$${t}=\sqrt{\mathrm{sin}\:{x}}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{\mathrm{sin}\:{x}}}{\mathrm{cos}\:{x}}{dt}=\frac{\mathrm{2}{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}{dt} \\ $$$$\Rightarrow\:\mathrm{2}\int\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}\:\mathrm{and}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{this} \\ $$

Commented by MJS last updated on 22/Mar/20

you should always test your solutions (d/dx)[2arcsin (√(sin x))]=((cos x)/(√(sin x −sin^2 x)))≠(1/(√(sin x)))

$$\mathrm{you}\:\mathrm{should}\:{always}\:\mathrm{test}\:\mathrm{your}\:\mathrm{solutions} \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{2arcsin}\:\sqrt{\mathrm{sin}\:{x}}\right]=\frac{\mathrm{cos}\:{x}}{\sqrt{\mathrm{sin}\:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}}\neq\frac{\mathrm{1}}{\sqrt{\mathrm{sin}\:{x}}} \\ $$

Commented by jagoll last updated on 22/Mar/20

so what the correct answer sir?

$$\mathrm{so}\:\mathrm{what}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}\:\mathrm{sir}? \\ $$

Commented by MJS last updated on 22/Mar/20

I cannot solve it there′s a solution using Γ(x) somehow, it had been on this forum, maybe a year ago, I can′t find it

$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{using}\:\Gamma\left({x}\right)\:\mathrm{somehow},\:\mathrm{it}\:\mathrm{had} \\ $$$$\mathrm{been}\:\mathrm{on}\:\mathrm{this}\:\mathrm{forum},\:\mathrm{maybe}\:\mathrm{a}\:\mathrm{year}\:\mathrm{ago},\:\mathrm{I}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{find}\:\mathrm{it} \\ $$

Commented by jagoll last updated on 22/Mar/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 22/Mar/20

found this: ∫(dx/(√(sin x)))= [t=x−(π/2) → dx=dt] ∫(dt/(√(cos t)))=∫(dt/(√(1−2sin^2 (t/2))))= =F ((t/2)∣2) =F ((x/2)−(π/4)∣2) +C it′s an eliptic integral but I cannot explain, it′s not my solution

$$\mathrm{found}\:\mathrm{this}: \\ $$$$\int\frac{{dx}}{\sqrt{\mathrm{sin}\:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}−\frac{\pi}{\mathrm{2}}\:\rightarrow\:{dx}={dt}\right] \\ $$$$\int\frac{{dt}}{\sqrt{\mathrm{cos}\:{t}}}=\int\frac{{dt}}{\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}}}= \\ $$$$=\mathrm{F}\:\left(\frac{{t}}{\mathrm{2}}\mid\mathrm{2}\right)\:=\mathrm{F}\:\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right)\:+{C} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{an}\:\mathrm{eliptic}\:\mathrm{integral} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{explain},\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{my}\:\mathrm{solution} \\ $$

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